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TPS561208: Figure15 TPS561208 VOUT = 3.3 V Efficiency, L = 3.3 µH

Hirotaka Matsumoto
Guru 27585 points
Part Number: TPS561208

Hi all

Would you mind if we ask TPS561208?

We would like to confirm Figure15 TPS561208 VOUT = 3.3 V Efficiency, L = 3.3 µH.

<Question1>
Could you let us know the detail infomration of the inductor? Maker, part number. 

<Question2>
Our customer measured the efficiency with two inductor coditions as follows.
However, the result was 74%.

4.7µH
Part No : VLCF4024T-4R7N1R4-2
DCR : 87mΩ

3.3µH
Part No : CLF6045NIT-3R3N
DCR : 19mΩ

<The result>
Vin=5.0V/106.5mA
Vout=3.3V/117.6mA
->The efficiency was 74%.

So, compared with Figure15, there is 16% difference.
If you have the advice to improve the efficiency, could you let us know?

Kind regards,

Hirotaka Matsumoto

  • 0
    TI__Genius 10070 points
    Hi,

    From the data you attached, I think it is not accurate, you need to measure the input voltage and output voltage on the input and output capacitors, there is power lost on the wires, if the voltage of power supply is 5V, the input voltage on the board is lower than 5V because of the loss of wires.

    Below is the test result with the inductor (APS07H30M3R3).
    Vin=4.9984V, Iin=0.0714096A, Vout=3.332848V Iout=0.09997501A, Efficiency=93.35%.

    Shawn
  • 0 in reply to Shawn Nie
    Guru 27585 points

    Shawn san

    Thank you for your reply.
    OK, we will inform our customer it.

    Kind regards,

    Hirotaka Matsumoto